How do you factor #9x^2 - 8y^2#?

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Answer 1 · 2015-07-13T00:21:33

#9x^2-8y^2 = (3x-2sqrt(2)y)(3x+2sqrt(2)y)#

This is a difference of squares, but only with the help of irrational coefficients.

#9x^2-8y^2 = (3x)^2 - (sqrt(8)y)^2 = (3x-sqrt(8)y)(3x+sqrt(8)y)#

using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=3x# and #b=sqrt(8)y#

Note further that #sqrt(8) = sqrt(4*2) = sqrt(4)sqrt(2) = 2sqrt(2)#

So we can express the factorisation as

#9x^2-8y^2 = (3x-2sqrt(2)y)(3x+2sqrt(2)y)#